Sarthaks Test
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The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

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Given:

a24 = 2a10

a +23d = 2(a +9d)

5d = a (i)

To Prove:

a72 = 4a15

Proof:

L.H.S. = a72

= a +71d

= 5d +71d [from (i)]

= 76d

R.H.S = 4a15

= 4(a +14d)

= 4a +56d

= 4(5d) +56d [from (i)]

= 20d +56d

= 76d

Since, L.H.S = R.H.S.

Hence proved

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